Question

cmr \(\dfrac{b^6}{a^2}+\dfrac{a^6}{b^2}\ge a^4+b^4\)

Answer 1

Sử dụng AM-GM, ta có:

\(\dfrac{b^6}{a^2}+a^2b^2\ge2b^4;\dfrac{a^6}{b^2}+a^2b^2\ge2a^4\)

\(\Rightarrow\dfrac{b^6}{a^2}+\dfrac{a^6}{b^2}\ge2\left(a^4+b^4\right)-2a^2b^2\)

Ta chứng minh \(2\left(a^4+b^4\right)-2a^2b^2\ge a^4+b^4\Leftrightarrow a^4+b^4-2a^2b^2\ge0\Leftrightarrow\left(a^2-b^2\right)^2\ge0\)

Vậy ta có điều phải chứng minh